text:
Why do modern aircraft tend to have angular tails?
In the WWII-era and earlier, many (most?) planes had rounded tails, even high-speed and high-performance fighters such as the Bf 109 K, Ta 152 (though it is more square-ish), P-38, Hawker Typhoon, and even the F-80. A notable exception to this trend is the P-51.

Modern aircraft, on the other hand, tend to have very angular tails. High speed fighters, to airliners, even relatively slow aircraft like the Cessna 152 or Piper Archer have very angular tails.

What is the reason for this change in trend?

You say the P-51 is an "exception" but it looks to have a very rounded tail to me. It also seems like older passenger planes (e.g.  DC-3, Lockheed Electra and Constellation) had rounded empennages as well, so I'm curious as to what caused the change. The fact that slow GA airplanes also have an angular empennage indicates it isn't just a speed-related design.

I guess we've found out later that rounding the tip is pointless.

@zymhan I suppose where you draw the line between "round" and "angular" is subjective to some extent, as it could be argued that the Ta 152's tail is angular with rounded corners, or round with straight edges. While the P-51's tail has rounded corners, I'm more referring to the overall shape versus corners. The Spitfire and P-38 have almost oval tails, while the P-51 has very distinct angles.

@flyingfisch Oh you're right, the last plane you listed was P-51, and then I saw the image of the P-38 and had a brain fart. Indeed, the P-51 has a much squarer/more angular tail.

Mainly looks perhaps?

@quietflyer why do high-speed fighter aircraft seem to be the first to adopt the angular tail? I doubt their design decisions were based primarily around looks...

@quietflyer It's almost definitely not looks.

@user3528438 care to elaborate? what was the perceived benefit to rounding the tail that was later found to be pointless?

@verandaguy I recall an anecdote about the C172 vertical stabilizer being swept for the looks, albeit after calculating that the efficiency penalty versus an unswept design would be small. Looks sell aircraft as much as performance; especially to low-hour, high-income pilots.

@AEheresupportsMonica I think that makes sense for GA aircraft, but again, I doubt the F-86 was given an angular tail just because it looked cooler.

@flyingfisch I'm saying there might have never been any benefit to begin with.

@user3528438 if that's the case, I wonder why it took so long for common aircraft designs to adopt the more angular tail. You'd think there would be some aircraft with rounded tails and some with angular tails in the era from 1903 to 1940. In addition I could be completely off-base but it seems like it's more difficult to mass-produce a round tail than an angled one?

@flyingfisch Because they didn't have computers?

Dupe of this: https://aviation.stackexchange.com/questions/65169/why-is-the-tail-group-of-virtually-every-airplane-swept-instead-of-straight/65172#65172

@verandaguy on low speed airplanes yes it is mostly for looks.  Cessna switched their whole line except the 180-185 to swept tails in the mid 60s, basically for styling reasons.  There was no performance benefit significant enough to justify the massive retooling required.  Nobody sweeps horizontal tails of light a/c (the Aerostar is one exception) because nobody cares about the look from below.

@JohnK I don't think it's an exact duplicate. I'm not referring to swept tails in general, the bf 109 and P-80 both have swept, but round, tails. I'm referring to the move from a rounded tail shape to a sharply angled tail shape

The rounded or elliptical design was found to be the cleanest aerodynamically in non-compressible subsonic airflow, which is why it's very common on aircraft of that era.  While the design is very slippery for speeds of Mach < 0.6 or so, it is more difficult to manufacture, which is why, near the end of the war, Spitfires had the 'clipped' wingtips as opposed to the original pure elliptical planforms.
The modern low aspect, sweptback, trapezoidal tail design is a better choice for higher airspeeds and supersonic flight ie Mach > 0.8 and mitigating flight control flutter due to compressibility effects, the same reasons we use sweptback wing.  Note as well that modern fighters use the same kind of planform for their wings and tail planes as they offer a compromise between acceptable supersonic aerodynamics and good trans sonic maneuverability.

" While the design is very slippery for speeds of Mach < 0.6 or so, it is more difficult to manufacture, which is why, near the end of the war, Spitfires had the 'clipped' wingtips as opposed to the original pure elliptical planforms."
I thought the wings were clipped to increase roll rate and low altitude airspeed?
https://en.wikipedia.org/wiki/Supermarine_Spitfire_(early_Merlin-powered_variants)

They may have additional benefits but one undeniable features a clip wings is it simply cheaper and easier to manufacture the things.  When you’re a nation fighting for its survival against a hostile power, and needs as many fighters as it can get its hands on, that tends to win out.

do you have a source on that being the reason? It seems odd that the RAF would start clipping wings to save costs once they were already winning the war

Against this explanation, no modern glider has elliptical surfaces.

The wings were clipped for the same reason Reno air racer wings were clipped, to make them go faster at low alt and improve roll rate, but mostly go faster.  Conversely, for the ultra high altitude interceptor version of the Spit, they made the elliptical tips longer.

Elliptical wing planforms have the lowest coefficients of drag out of any planform for subsonic applications.  Source: Elger, D., Engineering Fluid Mechanics.

As to their use in manned gliders, one can speculate on that, though there are plenty of scale gliders which make use of elliptical planform wings.  It is reasonable to suspect that a rectangular or tapered wing capped with winglets is more aerodynamically efficient and easier to build that an elliptical planform on large scale.

@John K and they're lighter and cheaper to build, too.  Just at the expense of higher approach speeds and degraded slow speed handling.

@flyingfisch the smaller wing span increases the roll rate and reduces mass (thus increasing speed), whether the wing tips are square or elliptical is largely irrelevant for that.

@MartinArgerami Modern gliders and indeed planes such as the 747 and 777 and A360 achieve elliptical lift distribution by varying the angle of attack from wing root to wing tips - this is more commonly known as washout. This achieves the same effect as an elliptical shaped wing but is far cheaper and easier to manufacture

@CarloFelicione Actually, wings with elliptical lift distribution have lowest drag coefficients. This can be achieved 3 ways - make the planform elliptical, vary angle of attack from root to tip so that when you graph the lift distribution it is elliptical or change the airfoil from root to tip so the lift distribution is elliptical. The planform solution was the simplest to get right when you don't have CAD. With CAD or just a lot of math you can do it without making the planform elliptical

And considering RJ Mitchell and his team only had slide rules available to them, and minus another 85 years of aerodynamic research, an elliptical platform was an easy solution here.

@JohnK the test results quoted here seem to suggest that the speed increase was marginal but the roll rate increase was massive (at least at low speeds, 105 deg/sec to 175 deg/sec @ 200 mph): https://www.quora.com/Why-did-some-Spitfires-have-clipped-wings-in-WW2

Aside from any subtle aerodynamic benefits of elliptical surfaces vs straight tapered ones, the move away from rounded profiles was mostly for ease of manufacture. A rounded profile has much more complex lofting demands, especially for the stabilizer part, and is more expensive to produce.  Properly done, there are going to have to be compound curves, requiring skins to be stamped in 3-dimensional dies, which has to be done in an annealed state and heat treated after, with rib flanges bent in mating contours.  Compare that to a straight edge surface that only needs skins with simple 2 dimensional bends that can usually be done in the final heat treat state (for larger radius bends).
Some light aircraft, like the Cessna 170, were able to get rounded looking profiles with mostly flat skins by using just a compound curve formed leading edge skin.  But even here going to a straight tail eliminated that requirement and a stabilizer could be skinned with a single wrap instead of a separate upper and lower or left and right skins and a die stamped compound curve leading edge skin.  As well, think of the reduction in rivet count and labour hours in general.
Over time the straight edged shape also came to be associated with modern, so "modern looking" and "cheaper" in combination become pretty hard to beat and round surfaces disappear except in special cases.

While discussing the lower-speed GA part of the equation (the original question appears to be not concerned with those types of planes, but it's still relevant to mention them, as your answer does), it's worth pointing out that marketing has its influence as well: Cessna models like the 172 actually had almost-square tail shapes early on, probably closer in efficiency to an elliptical shape, but later were changed to a swept shape, for no apparent reason other than so it "looks faster".

@PeterDuniho yes I've been saying that in other posts about swept tails. On low speed airplanes it's mostly for looks.  Which is why the horizontal tail is rarely swept; it's the side profile that matters.  In fact if anything, sweeping the rudder hinge imparts a slight nose up pitching moment when the rudder is displaced, although it's too subtle to notice.

Yes, in the case of light aircraft, backward slanting tails were the direction of marketing departments, not aeronautical engineers.  The reason they are not good is that in low speed, high angle of attack such as the pre-landing flair, the rudder is actually increasingly canted out of the wind.  It's rearward slant adds to the angle of attack to make the projection of the rudder's area in in the direction of flight smaller.  One counter example is the Mooney M20 series series.  These aircraft have vertical leading edges on the vertical stabilizer, and the rudder actually slopes forward.  This gives the rudder a larger projection in the landing flair, resulting in it having increasing effectiveness.  Whether or not it makes the Mooney M-20 series of aircraft look faster is irrelevant.  Mooney airframes are noted for low drag, resulting in high speed at lower fuel consumption compared to most other light aircraft designs.


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Is there a minimum altitude for ejection seats?
Is there generally a minimum altitude for ejecting in order for the sequence to function reliably (all the way to opening the chute, and slowing down to a survivable descent rate)? I've seen ejector seats test being done on the ground, don't know if there were actually test pilots in those seats though.

I assume all bets are off if you're not wings level (obviously a low-level inverted ejection has a 0% chance of success).

Not an answer, but some info: https://en.wikipedia.org/wiki/Ejection_seat#Zero-zero_ejection_seat

In 1971 an Air Force Pitol was flying so very low over 1st Marine Division some seven miles north of Danang Air Force Base. The pilot's control stick in his Fighter Jet had malfunctioned, and he was doing everything possible to keep crashing into Marine barracks. I heard this "very load noise" and ran out of a building just as the jet went overhead. The letters on the plane could be easily read. Suddenly, the pilot observed were the high Division ridges sloped downward to sea-level like a "V", and he was able to point the nose of his plane into that "V" and ejected. The plane crashed and explo

Let's hope for humanity's sake that there was indeed no "Test" Pilot...lol

@Prodnegel: Actually, for at least one zero-zero seat, there was!

It depends on the ejection seat.
Modern zero-zero ejection seats are designed to be usable at zero speed, zero altitude.
Older ejection seats had different minimums which varied from model to model.
On a side note, the early F-104 ejection seats required quite a bit of altitude to deploy properly, since the pilot exited through the bottom of the aircraft. This was done because there was some concern that the pilot would hit the F-104's T-tail with a standard ejection seat.
EDIT
It appears that it may actually be possible to eject from an inverted aircraft. According to Wikipedia (bold added for emphasis):

The minimal ejection altitude for ACES II seat in inverted flight is
  about 140 feet (43 m) above ground level at 150 KIAS. While the
  Russian counterpart - K-36DM has the minimal ejection altitude from
  inverted flight of 100 feet (30 m) AGL. When an aircraft is equipped
  with the Zvezda K-36DM ejection seat and the pilot is wearing the
  КО-15 protective gear, he is able to eject at airspeeds from 0 to
  1,400 kilometres per hour (870 mph) and altitudes of 0 to 25 km (16 mi
  or about 82,000 ft). The K-36DM ejection seat features drag chutes and
  a small shield that rises between the pilot's legs to deflect air
  around the pilot.

I read somewhere that some ejection seats require a minimum speed as well, do you know why that is?

@roe: not sure, good question. Maybe another answerer can help on that.

@roe: The minimum speed is for the parachute, not for the seat itself.  I only have this second-hand, though, from a pilot quoting his instructor in an article about something else.

Modern ejection seats like the ones capable of inverted flight will "right" the pilot after exiting the aircraft, then fire rockets to gain enough altitude. That's how you can eject 100 feet above the ground upside-down and survive

Pilots have successfully ejected at negative altitudes in a handful of instances, after being forced to ditch in water.
Documented evidence exists that pilots of the US and Indian Navies have performed this feat.
Also: I once read, but cannot now locate, the citation awarded a pilot of the USAF for the world record lowest altitude successful (i.e. the pilot survived) emergency ejection from a stricken aircraft over land: sixteen feet below ground level.  Apparently the young pilot was practicing with his T-33 single-engine trainer when the engine flamed out and could not be restarted.  The pilot observed the flat desert below, crossed by several paved roads and highways.  One road seemed to be completely clear of traffic, so he opted for a gear-down dead-stick landing on it, hoping to minimize damage to the aircraft.  He discovered only after touchdown that the paved strip was not a road at all but a flash flood drainage ditch, wide enough for the fuselage but not the wings.  The aircraft was seriously damaged and set on fire.  The damage prevented the canopy from opening, so the pilot used the ejection seat to escape burning to death.  He survived with minor burns, some compression damage to his spine, and two broken ankles.

The interpretaion of "negative altitudes" into "underwater" is a bit mistakable. You could also eject in Death Valley and be below MSL...

Actually I was interpreting "underwater" as "negative altitude", not the other way round.

Now that you mention it, I think Death Valley would be a perfect place to set up an ejection seat research and test facility -- just for the "negative altitude" bragging rights.

Yeah, I read the Wikipedia entry and kind of reverse-engineered your statement. So the article is kind of ambiguous. Since the incidents are Navy, I´d guess they ejected from a sinking jet.

Shouldn't "negative altitude", to have a useful meaning, be relative to ground, not sea level?

@Alexander Kosubek: For takeoff, landing, and terrain avoidance you want height above ground. For tactical intercept vector or flight path separation, you want altitude above sea level. For engine and turbocharger settings, cabin pressurization, and service ceiling, you want density altitude.

Quite a few years back, about 28 to be exact I worked on Mirage III service for the RAAF and one of my jobs was to fit the ejection seats after major airframe servicing.
Originally they were fitted with Martin Baker OM6 and these seats were zero-90 classification. They were powered by cartridges, basically a big multi stage canon that shot you out of the plane.
This means they were able to be used at zero altitude and 90 knots forward speed, the speed was purely to have enough air flow to open the chute fully but mostly to help with the removal of the canopy by blowing it backwards. The canopy jettison would unlock the canopy and lift the front into the airstream, the forward motion would do the rest.
Later we made a modification to the seats to make them into zero/zero capability and this included fitting a rocket pack under the seat that would launch the seat to a safe height enough for the parachute to open normally.
In both cases the seats were capable of going through the canopy but normally the canopy would eject first except in the zero zero situation where the seat would go through the canopy.

The Martin Baker MkGRU7 and EA7 on both the A6 Intruder and on the EA6B Prowler aircraft have "zero-zero" capability, which means that the seats could be fired at zero AGL and zero airspeed.

It all depends upon the seat type, altitude, speed, attitude and orientation of the aircraft during the ejection process.  Ejection seats just provide a velocity vector in order to give the pilot enough time to clear the crippled aircraft and deploy the recovery parachute, as this USAF training film shows.

This results in an envelope of airspeed, altitude, and attitude in which the seat can be activated allowing for a safe recovery.
For a so called 'Zero-Zero' ejection seat, it allows for a safe ejection from an aircraft on the ground in level flight attitude.  It does NOT guarantee safe ejection from an aircraft in unusual or inverted attitudes at low altitudes or airspeeds.
In the Boeing ACES II ejection seat a basic guide to seat characteristics is given here.
I can't comment on the capabilities of the Russian seats.  However the Russians do have a tendancy to be excessivly boastful and colorful with a gullible western press, such as the 1989 crash of a MiG-29 at the Paris Airshow, where the press ran with a story about the K36 seats being able to safely eject at low altitudes and extreme attitudes.  The reality was the seat in this incident was activated outside of its safety envelope; the pilot was just lucky he wasn't killed.

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Do lighter-than-air aircraft communicate with ATC?
Do lighter-than-air aircraft communicate with ATC? If not, how do they avoid running into other aircraft?

Short Version:
No aircraft is required to talk to ATC, except in certain situations.  Whether or not they have a radio, all pilots are required to continuously watch outside the airplane for other aircraft and to avoid them visually, much like someone driving a car avoids other cars by looking for them.

Long Version:
Aircraft are required to communicate with ATC only in certain situations.  14 CFR 91 contains the requirements:

91.126 - Within 4 miles and 2,500 ft of an airport in Class G Airspace
91.127 - Within 4 miles and 2,500 ft of an airport in Class E Airspace
91.129 - Within Class D Airspace
91.130 - Within Class C Airspace
91.131 - Within Class B Airspace
91.135 - Within Class A Airspace
91.155 - In weather conditions less than VFR
91.183 - While operating IFR
91.511 - More than 1 hour flight time or 100 miles from shore (overwater)

Many of these can be  waived with prior coordination with ATC.
Outside of the above situations, no radio is required.  In fact, there are still many airplanes that don't even have an electrical system, much less a radio, and they are perfectly legal to go fly around.
On the other hand, if they do have a radio (even a hand-held radio) it can greatly add to the safety of flight if they remain in contact with ATC since they can point out traffic that they might not have seen as quickly otherwise.
Now, whether or not the pilot is talking to ATC, they are required to "see and avoid" other aircraft, and certain aircraft have the right of way over other aircraft.  As you can imagine, an unpowered balloon can't really get out of way of a 747, so it is the responsibility of the more maneuverable aircraft to turn to avoid hitting the less maneuverable one (see 91.113 for more details).
They are also required to remain a certain distance away from clouds (91.155).  This is so that one aircraft doesn't come out of a cloud, possibly at a very high speed, and run right into another aircraft because they didn't have time to avoid them.  If they are far enough away, then the pilots will have time to see and avoid each other.

I just took a balloon flight in uncontrolled but incredibly busy airspace. The pilot had a handheld and coordinated with other pilots, both in balloons and in ground support vehicles. I imagine its similar anywhere there is congestion.

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What are the spiral marks in the center of the engines?
When I get a chance to peek inside an engine, I noticed that many have spiral marks painted in the center. What is the purpose of these?
Pictures:

It is supposed to deter bird strikes.

The benefit of these white markings inside the engine is twofold:

The ground crew can tell if the engine is running or not.
When an airplane is at the gate before departure, there are several personal attending the airplane. They would be loading luggage, refueling, removing waste products, loading food, etc. etc.. Not all of these people are in constant contact with the pilots and they can come very close to the engines. Although there are several precautionary measures taken before the pilots turn on the engines, it is a very clear visual indication by looking at the spirals to tell if the engines are running or not. Rotor or prop blades sometimes have their tips painted for the same reason.
The birds keep away.
During flight, when the engine is running, birds see the spinning spiral as opposed to just a black hole, and hence they stay away from it. It is like an anti-bird signal because if there was nothing in the middle it just looks like a cave or black void that birds tend to fly toward. Even though it looks like a blur to us, birds can see motion better. Wikipedia mentions this also:

... it appears as a white circle which discourages birds from flying into the engine.

Rolls-Royce brochure (PDF) quoted the following (identified by user BowlOfRed in comment below):

Our aerospace engines have swirls painted onto their spinners in order
  to indicate when the engine is rotating while on the ground. In flight
  these swirls flicker as the engine rotates at high speed, scaring
  birds and allowing them to fly clear of the engine.

I'm skeptical about the bird deterrent.

Couldn't find it on the live site, but this archival Rolls-Royce brochure agrees with the bird statement.  https://web.archive.org/web/20061019190934/http://100.rolls-royce.com/files/852/facts_100.pdf  (#10 on list)

I too was "told" that it scares birds away, but I've yet to come across any scientific evidence supporting the claim.

They may help make the engines appear as eye-spots tot the birds in front. Worth researching I think.

@kevin  For the purposes of answering the question, it doesn't matter if it works or not.  The fact that Rolls Royce believes it works would explain why they put it on there (even if that belief is erroneous).

@fooot You could try asking this on the Skeptics site.

I've seen ground staff inspecting the fans during turnaround.  I'm sure that they appreciate the unambiguous display of fan movement.

In this video a Lufthansa pilot performs a preflight inspection and makes the claim about it being a bird deterrent.

It is possible that it could deter birds while the engine was running on the ground during taxi operations.  During flight or on the takeoff/landing roll, I think that, by the time the aircraft got close enough for the bird to make out the flickering engine inlet, it would already be too late for the poor creature to avoid a birdstrike.

As a long time ramp agent for Delta and Southwest Airlines, I can tell you they are a critically necessary element to ground crew safety. Not only do they tell you the impeller in spinning, but once the human eye can no longer identify the swirl pattern, it is spinning with enough speed to be an induction hazard (it can suck you in).
When parking an airliner, the crew lead guides the aircraft using aircraft marshaling wands. When he/she turns both tip toward each other, the crew set chocks around the wheels. Then NOTHING happens until that crew lead can see those spinning spirals clearly in all engines. Then and ONLY then, does that lead clear the crew to continue operations and begins comms with the flight crew.

The spiral on many GE engines is a stylized "G":

Source
Rolls Royce has a longer spiral.

The design helps with ground crew safety, as Farhan's answer discusses. However, it is not required and not all engines have them, suggesting that they are also a matter of style between engines and manufacturers.
This page from Boeing about bird strikes says that there is no scientific evidence that spinner markings, airplane colors, landing lights, or weather radar deter birds. On the other hand, the Museum of Flight states that "research has shown" the spiral deters birds. Though I suppose this research could have been unscientific.
The AOPA says that "one air carrier detected slightly reduced bird strike rates after painting the jet engine spinners white." If there is indeed an effect here, it would seem that a spiral would be less effective than painting them white.
This paper notes mixed results on using eye spots on aircraft. When used on the ground, the birds seem to get used to them quickly, just as they do with the noise.
As costly as bird strikes can be, it would seem that aircraft operators would widely adopt anything shown to help prevent them.


text:
What are the various fees associated with flying?
I am aware of airport landing fees, fuel fees etc, but it was brought to my attention that there is airspace charges? And especially in Canada, the fees are horrendous. What are the various types of fees airlines and private planes have to pay in terms of airspace and air traffic control fee around the world, and Canada and the US?

Possible duplicate of How does a national ATC make money?

I don't think it is a duplicate. This question is asking what are the various fees (collected by third parties) that a pilot or operator should be aware of.

@kevin He is asking specifically about fees related to airspace and air traffic control.

I also don't think it's a duplicate.  I think it's too broad.

I will answer this for the US as that is the jurisdiction I know best.
There are no airspace or ATC usage fees here in the US. The airspace and its resources are free for any plane of any size that conforms to the FAA's FAR's and is legal to fly in a given airspace.
The FAA is in the process of adding overflight charges for aircraft that fly through our airspace but neither land nor depart from the US.
The only fee (somewhat indirectly related but not a component of the airspace its self) is for planes coming in from abroad who will need to pay some customs fees to enter the country.
Strictly speaking you may be required to carry IFR charts or VFR charts on board to conduct the flight in question in US airspace. These must be purchased from an official source. The FAA maintains them but the Pilot must pay for them. Until recently the FAA also physically printed them and they were purchased directly from them but this is no longer the case.
If you extend the question to include airports the answer will drastically change for the US.

Overflight charges would count wouldn't they?

Interesting, I did not know about those, adding to the answer now.

I didn't either until this question, but it seems that it is quite common for countries to charge those that fly over but do not take off or land in that country. Well, at least Canada and U.S.

Don't some (larger) airports charge landing fees?

Almost all do but their question was about ATC and airspace if they include airports I will extend my answer.

You are not required to have charts on board for VFR operations. For IFR, you can use EFBs and are not required to have any paper charts.

It is worth noting that aeronautical charts are available digitally and can be printed at home without any direct fees incurred. And by way of correction, there is no requirement for charts to be purchased from an official source.

I do not believe that charts printed at home are legal for navigation. But yo u are correct they are available digitally.

The FAA has an "Airways and Aviation Trust Fund" which (in essence) provides a large chunk of the FAA's funding. The AATF is itself funded in large part from a surcharge on aviation fuel, so that is a kind of fee charged on aviators who buy fuel in the US.


text:
Why aren't airliners spin-tested?
According to Wikipedia, only single-engine aircraft with an MTOW no greater than 12500 pounds are required to demonstrate the ability to recover from a spin, and not large or multi-engine aircraft.  This seems backwards to me, as it would seem to be far more critical for an airliner with potentially hundreds of people on board to be able to successfully escape from a spin than it would be for a small GA aircraft where only the pilot's life is at stake, and not anyone else's.
Can someone explain this seeming contradiction?

Little single engine airplanes have stall spin accidents maneuvering to land all the time because of the way they are used.  When is the last time an airliner got into a spin on approach?   You have to cater to scenarios with a certain level of probability, not every possible thing that could ever happen.

1. Spins are less frequent on large aircraft

They are less maneuverable
They are flown by more experienced pilots
They have more protection systems (e.g. flight envelope protection, yaw damper) to avoid uncontrolled flight

2. Spins may cause structural damage on large aircraft
The larger the aircraft, the lower its tolerance to stress. While one can quite comfortably spin a Cessna or Cherokee, attempt one on a Boeing 777 and things will break. Gear doors, the bolts holding the vertical stabilizer as well as cabin accessories are not designed to handle stress in that direction, not to mention engine flame-out.
3. Spins may not be recoverable
As discussed in this question.

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text:
Why are the center tank pumps deactivated when slats are extended on the A320?
Why are the center tank pumps deactivated when slats are extended on the A320? I researched about this and all I can see is hints. They say it is because of regulations or certifications.

"they" who? where did you read about this?

I read it at aviation forums my friend, they say it's some kind of certifications. I also read part 23 but it is not directly stated.

I believe it's an interpretation of an airworthiness requirement in 14 CFR 25.953:

Each fuel system must meet the requirements of § 25.903(b) by -
(a) Allowing the supply of fuel to each engine through a system independent of each part of the system supplying fuel to any other engine;

where 25.903(b) reads:

(b)Engine isolation. The powerplants must be arranged and isolated from each other to allow operation, in at least one configuration, so that the failure or malfunction of any engine, or of any system that can affect the engine, will not -
(1) Prevent the continued safe operation of the remaining engines; or
(2) Require immediate action by any crewmember for continued safe operation.

When the center tank is feeding the fuel on the (typical) A319/A320 feeds directly into the fuel manifold, acting as a single tank outlet for both engines (see also 14 CFR 23.953). Additionally, takeoff with crossfeed open is prohibited for the same reason.
The A321, newer A319/A320, and other designs like Boeings, feed fuel from the center into a wing tank first, avoiding this requirement. Indeed, if you have an A321 or the A319/A320 with the new system, the limitation is deleted.
It would not be for purposes of high fuel pressure, since when center tank feeding, all pumps run, just the center tank pumps overpower the wing tank pumps.

The 737 has similar sequencing valves built into the fuel pumps, the wing pump pressure relief is set lower than the center tank pumps. So there is no difference with the A319/A320. (By the way, the A321 has a center tank by default, you may be thinking of the additional center tank)

The center tank pumps transfer fuel directly into the engines. When the slats are extended, the pumps are disabled to ensure that the wing tanks feed the engines rather than the center tank. The reason for this is to account for a possible tank pump failure. The tank pumps are electrically powered so there is a tiny bit of a chance of an electrical failure which can take the power off them. Unlike the center tanks the wing tanks can be used to gravity feed the engines.
With gravity fuel feeding active, the aircraft engines can still be supplied with fuel with no push from the tank pumps. The aircraft computers use the slat extension as a cue that the aircraft is preparing for a take off. Without gravity feeding, multiple tank pump failures can mess up fuel flow into the engine. This is the last thing you want to happen close to the ground. The center tank pumps are disabled for this reason. To allow for gravity fuel feeding by the wing tanks.
When the A321 came out Airbus made its fuel system simpler by replacing the center tank pumps with jet pumps (works with motive fuel) and put two transfer valves in the center tanks. So, in the A321 the fuel is pumped from center tank into the wing tanks which then transfer fuel into the engines. So, the engines are always supplied by the wing tanks which takes away the need for the previously mentioned logic. The post 2014 A320 models have the A321 system in place. Thus, it does not apply to them either.

A320 fuel tanks pre mod and post mod.

Its to provide an independent fuel supply to each engines. If both engines are fed from the Centre and its contaminated then both engines fail. If both are being fed from their respective wings then the supplies are independent, that's all.

Welcome to Aviation.SE! Could you provide more details? This doesn't seem to add much to the existing answers.


text:
How does an engine failure affect the flight of an airplane?
I'm not even close to an aviation enthusiast, but after watching some videos I found I have a deep misunderstanding on how airplanes work.
.
I've seen many cases where one of the engines fail and the plane has to fly on a single engine for a long time.
How doesn't the fact that the forces are now uneven drive the airplane to a side drift? (I'm sure this is all wrong, but I just don't understand how)

@ymb1: Unless that plane's had a double engine failure, in which case the arrows are the right way around for the force generated by those giant windmilling drag buckets.

It does.  Asymmetric thrust will in fact force the airplane into a turn unless the pilot counteracts this by using rudder and aileron to command an equal turn force in the other direction.  Typically both opposing rudder, and banking away from the dead engine will be used to result in straight (although uncoordinated) flight.

oh, so I was right? I watched a video about Air Transat 236, and it didn't explained that. thanks.

Well, from 06:13 until 06:26, the plane was flying with one engine providing thrust. During those few minutes, the pilot would have had to do as Skip suggests. After 06:26, the second engine died due to fuel starvation and the airliner was then a glider.

@wingleader: The interesting part of that incident was when both engines failed. One engine failing happens all the time and is not serious trouble (unless there is a secondary damage) so they would not consider that phase worthy of detailed discussion.

@Jan One engine failing happens all the time? you couldn't scare me less ):

Jan has a point: See the Aviation Herald website for all kinds of events in aviation (http://avherald.com). The last plane with engine trouble was a 767-300 on May 22nd. Given the number of flights per day worldwide, the probability for a single airplane is rare, indeed.

Is an engine failing ever the kind of thing that can go unnoticed by the passengers, so that it could conceivably have happened to me? Or is it inevitably "we are sorry to announce that we're immediately diverting to the nearest airfield large enough to land, please try not to panic"?

Passengers would very likely notice. There'd be a change in sound, and the pilot would have to "feel around" for the right balance of controls to compensate for the asymmetrical thrust. And, yes, I'd expect a divert to the nearest suitable airfield  if the plane wasn't very close to its original destination.

@wingleader: All aircraft that carry more than 12 passengers must be able to fly with one engine inoperative (tested during certification) and pilots train it at least twice a year in simulator. There is really no immediate danger in such incidents (unless it's uncontained failure with secondary damage, but even Quantas 32, which was rather serious uncontained failure, flew for another almost 2 hours and safely landed).

@RussellBorogove: In 2-engine aircraft, it is always emergency (there is no danger yet, but there is no redundancy any more) and land at nearest suitable airport. In 4-engine aircraft the flight may be continued if there is enough fuel and no risk of further damage (usually when the engine was shut down manually because of indications outside limits rather than quit itself). Then it's up to discussion with dispatch.

@JanHudec: just for posterity because I have this discussion all the time: per definition, an engine failure in a perf class A is an urgency (pan-pan) not an emergency (mayday).

As a pilot, I really don't care if I have 10 engines on my aircraft. If one fails, it happened for a reason. And that reason can make the next one fail too. In my opinion, it's irresponsible of a pilot to continue flying after a failed engine - especially one that can't be restarted. It may not be an emergency (though I don't know how it's not), but it's darn sure urgent enough to head to my nearest alternate. As the Pilot In Command, I'd always prefer the safe decision. And I wouldn't say that it "happens all the time".

@Radu094 and what happens when your HYD pump on the good motor decides to quit?  Engines do a whole lot more than just provide thrusties; in fact, I'm much more concerned with the accessories I might have just lost with an engine failure, vice the actual engine itself.

It doesn't matter: Calling it a mayday instead of pan-pan is at the captain's discretion, but technically a mayday implies you are no longer capable of sustained flight and are crashing. This stuff should've been taught in Radio Comms classes for PPL

@Radu094: Mayday implies there is an imminent danger to the flight. It does not necessarily mean it's not capable of sustained flight at all.

@JanHudec: incapable of sustained flight is maybe stretching it a bit , but "imminent danger and requiring immediate assistance" usually means just that in aviation. In fact, you will have to work hard to argue the "imminent and immediate" requirement in an aircraft still capable of flight ( ahem and not on fire). Maybe this belongs in a chat...

@RussellBorogove: And even then, they'd still be going to the nearest suitable airport - it's just that the nearest suitable airport happens to also be the original destination.

@Vikki Happend to me once... LAX-HNL, engine failed juuuust before the midpoint. Return to LAX. Nobody noticed but me. "Excuse me, are we turning around? — No sir, why would we? — Well that's what I see on the flight map — Oh, I'll go ask the captain".

It's just like sticking your hand out the window to catch the wind while driving. But just in a airplane rather than a car, and rather than your hand it's the rudder - on massively different energy scales. Same concept, physics wise.
The side drift in your example is compensated by the rudder (aka. like on a ship, aka. the lefty-righty bit, aka. the Yaw Axis in Pitch/Yaw/Roll terminology)
See this in action on the other side of the fence, when engines are working perfectly, but the Earth decides it doesn't want to play ball in a storm, i.e. when winds are forcing planes to land at a 45° angle to the runway (Gibraltar is common, or Manchester Airport as it's just not that good, nothing to do with the weather or pilots etc. .....)  - All the airplane giblets (Pitch aka. up-y down-y / Yaw aka. lefty-righty / Roll aka. tilt-y-left-y tilt-y-right-y mechanisms)  are playing ball, but the external physical forces of the storm are just over powering the airplane capabilities - hence, things are a bit off kilter resulting in things going skew-whiff.
Commercially speaking, it basically costs a sh!t tonne more money to travel the same distance. I'm sure there is a mathematical calculation behind this based on physics. Please do let me know if you know how to calculate this in the comments!


text:
Difference between EASA and FAA regulations for airliners?
The design of large transport aircraft is governed by EASA's CS-25 regulation and the FAA's FAR Part 25 regulation.
It seems however that these regulations are nearly identical. It is obvious that a lot of collaboration took place between the two authorities in the creation of these standards.
Are there any major differences between these standards?

Unless there are just too many large differences, I don't think this question is too broad.

I know the regulations are massive documents and with a lot of sections and I'm sure there are a few small differences but am really interested, as the question states, in major differences.

@Jonny I'd make your intention to ask for the major differences only easier to recognize in the question. Formatting helps a lot here ;)

There are literally thousands of differences between them, including many major ones. This question is indeed too broad.

Unless this is converted to a community wiki, the question is too broad.

My motivation for the question was driven by the study of the Sections to do with emergency egress (See sections 25.803,807,809,810,811,812,813,815,817,819 and 25.820) These sections all have the exact same names and numbers in both regulations. I see emergency egress as a major part of the regulations and in this regard the regs are in nearly exact alignment. I understand that there will be many many minor differences but I truly don't believe that, from the perspective of aircraft design, there are so many major differences. After all, the aircraft are operated on a global stage and

surely need to adhere to the regulations of the countries in which they are serviced and operated.


text:
Is (L/D)max achieved at a specific speed or AOA?
I'm little bit confused about the (L/D)max.
So I've read this blog: https://joeclarksblog.com/?p=4155
and the author shows this graph:

So if I understand it correctly the (L/D)max is at the intersection of parasitical drag and lift-induced drag. Thus, at the point which has the minimum total amount of drag. Which is achieved at a specific speed.
Skybrary states the following:

The maximum lift/drag ratio occurs at one specific CL (Lift
  Coefficient) and AOA (Angle of Attack (AOA))

https://www.skybrary.aero/index.php/AP4ATCO_-_Lift/Drag_Ratio,_Forces_Interaction_and_Use
Here is is the lift coefficient:

Source: https://en.wikipedia.org/wiki/Lift_coefficient
The graph shows that the airfoil is the most efficient at a specific AOA. Thus in the ranges between 15° and 18°.
sciencedirect.com states the following:

The L/D is not constant for a given wing, however, but changes with
  the angle of attack. Researchers thus normally use the maximum
  lift-to-drag ratio, L/Dmax, to characterize a given wing.

https://www.sciencedirect.com/topics/earth-and-planetary-sciences/angle-of-attack

I'm confused because one source tells me that (L/D)max is achieved at a specific speed. The other source tells me that (L/D)max is influenced by the AOA.
Can you clarify this thought twist?

"Researchers thus normally use the maximum lift-to-drag ratio, L/Dmax, to characterize a given wing." Researches do this, pilots also look at "hw abrut is the stall" (fundamental for a nice flare), "how does this profile behaves when dirty/wet?", "what's the L/D when flying not at the exact best L/d AOA?", "How steep is the alpha-momentum curve?"

You're confusing maximum lift with maximum lift to drag ratio. Maximum lift is the largest lift force that an aerodynamic surface can provide at a specific speed (more accurately, dynamic pressure). Lift to drag ratio is a measure of how efficient the surface is at generating lift at the expense of drag (for an airplane, drag means more propulsion required).

Induced drag is dependent on Angle of Attack while parasitic drag is dependent on airspeed.  However there is a correlation between airspeed and angle of attack so the L/D graphs are usually given using airspeed.  The airspeed is also more useful to pilots than angle of attack as most airplanes don't have a AoA sensor.
So really both sources are correct here but from a technical standpoint L/D is mostly dependent on Angle of Attack although parasitic drag is completely dependent on airspeed.

Can you explain the correlation between airspeed and AOA?

@Julian:  This website explains it pretty well:  https://www.av8n.com/how/htm/aoa.html#sec-ias-aoa

I chose this answer as the best answer because the link clarifies the thought twist I had.

The induced drag also depends on airspeed, right? Don't both induced drag and parasitic drag vary as the square of the airspeed? (And then, in straight and level flight, angle of attack varies with airspeed, which makes everything confusing.)

I'm confused because one source tells me that (L/D)max is achieved at a specific speed. The other source tells me that (L/D)max is influenced by the AOA.

The speed is only correct for a specific wing loading and straight flight at 1g. Only then will the optimum lift coefficient be reached in level flight at that speed.
The angle of attack version is the more generic and better way. Actually, optimum L/D is achieved at a specific lift coefficient, which is closely related to the angle of attack via the lift curve slope.

The graph shows that the airfoil is the most efficient at a specific AOA. Thus in the ranges between 15° and 18°.

No, it only achieves the highest lift coefficient in that range. This is right before stall and certainly not the position of optimum L/D. Drag will have grown disproportionally over the previous angle of attack range. Without a drag polar it is impossible to say where that specific airfoil has its best L/D.

wing loading and unaccelerated flight

"Eta is geometric".
L/D is referred to the angle of attack.
This is way easier understood looking at gliders: a glider with e.g. a best L/D of 50 will, in calm air and when gliding at the AOA where such L/D is obtained, advance 50 miles for every mile of altitude lost.
The speed at which such gliding flight will happen is depending on the loading: the heavier it is, the faster the speed.  For this specific reason, water ballast are common in gliders: if the pilots wants to fly faster he can just "point the nose down" and follow a steeper path but in doing this the wing will be working at an AOA where the L/D is gets farther from the optimal one whilst adding mass the the aircraft he will be flying faster without having the wing working at AOAs with lower L/D.
Same concept applies to power aircraft but it's harder to mentally visualize hence the choice of a glider for this answer.
Often flight envelope characteristics such as stall speed best L/D speed etc are expressed well as.... speed for pilot's convenience and making a few assumptions (MTOW load and level flight for stall speed, unballasted weight of a glider for best L/D) but in reality there are no stall speed nor best L/D speed: there are stall AOA and best L/D AOA)
Let's look at a glider's polar and L/D, referred to the L/D measured on the whole aircraft and not only on a wind tested in a wind tunnel (but similar results would be expected thee):
L/D

Polar

images from second answer of this interesting question
As you can see the curves have the same shape but are shifted depending on the wing loading, especially in the first graph can be seen that the same best L/D of about 47 is obtained at a higher speed the higher the wing loading (and thus the total weight since they all refer to the same aircraft), that's the speed where the wing flying at its optimal AOA gives a lifting force equalling the total weight thus making for an unaccelerated descent.
As for what is causing the wing's polar to have that precise shape: yes, the various forms of drag enter into account and this would be an interesting question you should ask: why polars have their characteristic shape?

What is the vertical axis of the second plot?

@jimmy: sink speed in m/s

I would recommend adding explanation on why the second plot matters (for example, best glide/climb speed coincides with L/Dmax for that weight, and why). I also don't think the second plot could be called a polar. Good data.

@Jimmy -- "A polar curve is a graph which contrasts the sink rate of an aircraft (typically a glider) with its horizontal speed." https://en.wikipedia.org/wiki/Polar_curve_(aerodynamics)

@quietflyer In airplane design world, polar means something a little different, although it would share the exact same characteristic as the plot shown in the answer. I will defer the nomenclature to those of you who are more familiar.

@Jimmy -- yes I understand there is more than one kind of "polar" -- https://en.wikipedia.org/wiki/Drag_polar

An edit to my previous comment:
In airplane design world, polar means something a little different. The plot makes physical sense to me; I'm just recommending to the answerer to clarify how it relates to OP's question regarding AOA and airspeed vs. max L/D.
I will defer the nomenclature to those of you who are more familiar.

For any aerodynamic surface, lift and drag are:
$$L=qSC_L$$
$$D=qSC_D$$
where $q$ is dynamic pressure and $S$ is reference area.
Assuming a thin, high aspect ratio, non-swept lifting surface at low Mach, like the wing of a glider, we can divide its drag coefficient into two components, as the Wikipedia entry described: parasitic drag, $C_{D_0}$, which we can assume to be a constant, and induced drag, $C_{D_i}=KC_L^2$, where $K$ is a constant dependent on the shape of the surface (aspect ratio, taper, etc.).
At $L/D_{max}$,
$$\frac{d(L/D)}{dV}=\frac{d(C_L/C_D)}{dV}$$
$$=\frac{1}{C_{D_0}+KC_L^2}\frac{dC_L}{dV}-\frac{2KC_L^2}{(C_{D_0}+KC_L^2)^2}\frac{dC_L}{dV}$$
$$=1-\frac{2KC_L^2}{C_{D_0}+KC_L^2}=0$$
So,
$$C_{L_{L/D_{max}}}=\sqrt{\frac{C_{D_0}}{K}}$$
There is a unique AOA on the lift coefficient curve, like the one you presented, that corresponds to this $C_L$ (pre-stall), so the statement:

The maximum lift/drag ratio occurs at one specific CL (Lift Coefficient) and AOA (Angle of Attack (AOA))

is indeed correct and there is a unique $L/D_{max}$ for a given lifting surface (see the first plot presented by Caterpillaraoz, notice how remarkably similar the maximum $L/D$ is for different wing loadings).
This seems to directly contradict with your second source:

The L/D is not constant for a given wing, however, but changes with the angle of attack. Researchers thus normally use the maximum lift-to-drag ratio, L/Dmax, to characterize a given wing.

Notice I was only talking about lifting surfaces. For a whole aircraft, there is also trim drag, which is AOA dependent. And if we deviate from our assumption of thin, high aspect ratio, non-swept, low Mach, then $C_{D_0}$ is no longer a constant and becomes a function of Reynolds number, AOA, AOS and Mach. For high speed airplanes, or airplanes with flaps/slats, our simple model breaks down, and $L/D_{max}$ is indeed, AOA dependent.

+1 for specifying "Assuming a thin, high aspect ratio, non-swept lifting surface at low Mach". Simplifications are cool and telling the conditions where such simplifications are valid is even cooler.

"So if I understand it correctly the (L/D) max is at the intersection
  of parasitical drag and lift-induced drag. Thus, at the point which
  has the minimum total amount of drag. Which is achieved at a specific
  speed"

Actually, speaking in the most general terms, neither L/D max nor maximum Lift nor minimum Drag are achieved at a specific speed. The graph tying the L/D ratio to a specific airspeed is ASSUMING one specific weight, and also is ASSUMING 1-G flight. In many cases this is a perfectly reasonable assumption, but not always.  Fundamentally speaking, L/D ratio and lift coefficient and drag coefficient are all functions of angle-of-attack, NOT airspeed.
However, in 1-G flight at a given weight, there's a one-to-one correlation between angle-of-attack and airspeed.  Once you understand this, your confusion will vanish.
If we DO assume 1-G flight at one specific weight, then it's fine to say that variables like L/D ratio, lift coefficient, and drag coefficient are functions of angle-of-attack, or of airspeed, whichever we prefer.  When presenting the information to a pilot who will be interested in the aircraft's performance in 1-G flight at some given known or assumed weight, then obviously it makes more sense to present the information in terms of airspeed, because the pilot has an airspeed indicator but not an angle-of-attack indicator.  This would pertain more to L/D ratio than to lift coefficient or drag coefficient, because the L/D ratio is tied directly to something the pilot is actually interested in optimizing (the glide ratio), while the lift coefficient and drag coefficient aren't tied directly to any one parameter that the pilot cares about.  (The lift coefficient is NOT a measure of the "efficiency" of the wing, as the question suggests.)  Hence the reason you are seeing the L/D curve presented with airspeed along the X axis, rather than with angle-of-attack along the X axis.  It's a graph that's meant to tell the pilot something he can actually use in the cockpit, while flying in the 1-g condition at some known or assumed weight.
Here's another bit of trivia to muddy the waters:
If the L/D graph is drawn for horizontal powered flight, the point of max L/D will occur exactly at the point of minimum Drag (not the same as minimum drag coefficient), because in horizontal powered flight Lift = Weight, assuming no complications from a tilted thrust line, and Lift is therefore constant.
If the L/D graph is drawn for gliding flight, the point of max L/D will occur ALMOST at the point of minimum Drag, but not exactly, because in gliding flight Lift is not constant.  It is slightly less than Weight, and more so at poor glide ratios than at large glide ratios.  The difference will likely be too small to notice though-- for all practical purposes the gliding L/D curve and the horizontal L/D curve are interchangeable, except in aircraft with very poor glide ratios.  Therefore for all practical purposes, the L/D curve can also be viewed as a Drag curve (not the same as a drag coefficient curve) in gliding flight as well as in horizontal flight, except in aircraft with very poor glide ratios.
In both cases (gliding flight or horizontal flight) the point of max L/D will occur at the point of max (lift coefficient / drag coefficient), which will occur NEITHER at the point of max lift coefficient nor at the point of minimum drag coefficient.
(By the way, all the above statements are true regardless of whether we are putting angle-of-attack, or airspeed, along the x axis of our graph.)
Always keep in mind that Drag is not the same as drag coefficient, and Lift is not the same as lift coefficient.  However the Lift/ Drag ratio is always exactly equal to the ratio of (lift coefficient / drag coefficient).

-1: Max L/D has nothing to do with minimum drag (for low speed, min drag would occur at zero lift), or whether it's powered or unpowered, unless your thrust line is offset from your waterline CG, which would introduce some trim drag.

@Jimmy -- the whole point of the "polar curve" (airspeed versus L/D) well-beloved of glider pilots is that you are assuming 1-G steady-state flight.  Otherwise, there would be no way to assign an airspeed to any given angle-of-attack value, and all you could do would be to create a curve showing angle-of-attack versus L/D.  And obviously, zero lift is not compatible with any form of steady-state flight except a vertical dive.  A steady-state vertical dive may well yield the lowest drag COEFFICIENT, but it doesn't yield the lowest Drag VALUE, because the airspeed is very high, and the (ctd)

@Jimmy-- continuing-- A steady-state vertical dive may well yield the lowest drag COEFFICIENT, but it doesn't yield the lowest Drag VALUE, because the airspeed is very high, and the Drag force in pounds or whatever is related to both drag coefficient and AIRSPEED SQUARED.  So, yes, absolutely max L/D has a great deal to do with minimum Drag, as I explained in the answer.

@Jimmy-- actually I've slightly mis-spoken.  The graph of airspeed versus L/D is normally not called the "polar".  That word is usually used for the graph of airspeed versus sink rate.  Sorry about that.  "A polar curve is a graph which contrasts the sink rate of an aircraft (typically a glider) with its horizontal speed." https://en.wikipedia.org/wiki/Polar_curve_(aerodynamics)

@Jimmy -- so to sum up, in steady-state horizontal flight, assuming no complications from  a tilted thrust line, minimum Drag (as a force in pounds or Newtons or whatever) IS obtained at the point of the max L/D ratio.

I would, respectfully, disagree on every single point made in your first two comments. Airspeed has nothing to do with 1G flight or not, it is the speed at which air moves relative to the aircraft. As demonstrated in my answer, L/D is equivalent to CL/CD under simplifying assumptions, so they are readily relevant.
I would encourage you to plot AOA vs CD and AOA vs CL in the same graph. Then you will see where minimum drag is.

@Jimmy -- L/D is ALWAYS equal to Cl/Cd.  Everything else cancels out.  I think you are confusing minimum Cd with minimum Drag.

@Jimmy-- I'm not saying that airspeed has no meaning if you aren't in 1-G flight.  I'm saying that you can't assign an airspeed value to a given angle-of-attack unless you have certain constraints and you know what those constraints are.   For a given aircraft, one given angle-of-attack will yield max L/D or max Cl/Cd, whichever you prefer to call it.  But my  comments about minimum Drag occuring at max L/D are based on the assumption that Lift is constant, not varying, across the airspeed spectrum.  Which means that we have a constant G-load and also a constant glide angle.

@Jimmy -- " Which means that we have a constant G-load and also a constant glide angle."  -- 1-G horizontal flight being one particular subset of these conditions.

@Jimmy -- OK enough, I need to go now. I'm surprised this hasn't already been booted out to chat.

I understand your answer a lot more now. Yes, I mistook CD with total drag. I would remove the down vote, if it allows me?

Thnks, it will allow I believe.  Only doesn't allow when quite some time has passed since vote, with no edit to answer

Let us continue this discussion in chat.

After reading and digesting all the answers I have a bigger understanding of the drag curves graph.
In short, the displayed drag curves graph applies for a specific weight in a 1-G flight. This graph uses the speed for the x-axis, because it's more useful for the pilot in the cockpit. It's new for me that it is possible to substitute the AOA with speed, and using speed as the x-axis (instead of the AOA) in the drag curves graph.
So there is still much to learn for me. +1 for taking the burden of writing this long answer but useful answer.


text:
What would be the ground roll and total distance to clear a 50ft obstacle given these conditions?
I'm a student pilot, the question I am trying to answer is

What would be the ground roll and total distance to clear a 50ft obstacle given an elevation of 5,000ft, altimeter 29.52, and a temperature of 25° Celsius, winds calm.

Disclaimer: I'm going to double check with my CFI, obviously, you should too if you are unsure.

So I understand how the chart works, I'm just unclear about the "Elevation 5,000ft. Altimeter at 29.52".
My inclination is to assume that the meaning is sea level pressure is 29.52, and therefore we should adjust our input for "PRESS ALT FT" to be the altitude that reflects that average pressure, which would be 500 feet.
We round up for safety, so assume PRESS ALT=1000' and TEMP=30° Celsius, we would have a ground roll of 890' and a takeoff distance of 1645', right?
Another question: Where do I find the official documentation for converting barometric pressure into the altitude that has that average pressure?

We round up for safety, so assume PRESS ALT=1000' and TEMP=30°
Celsius, we would have a ground roll of 890' and a takeoff distance of
1645', right?

Good thinking, but no.
Refer to the Pilot's Handbook of Aeronautical Knowledge, Chapter 10.  You want page 10-3 specifically.
When the altimeter setting is 29.92, the pressure altitude is the same as the field elevation.  When the altimeter setting changes, you must apply a conversion factor to the field elevation to get pressure altitude.
In this case, the altimeter setting is 29.52, which means that your conversion factor will be about 380 feet:

Therefore, the actual pressure altitude will be 5,380 feet.
If you round up for safety, the takeoff distances will be 1455 and 2855.

We can get a little more precise than that, though.
To calculate the ground roll at 25 celsius, we can interpolate between 20 and 30 degrees:
5,000 feet
$\left[ \left(1315 - 1215\right) \over 2 \right] + 1215 = 1265$ ground roll at 5000
$\left[ \left(2525 - 2320\right) \over 2 \right] + 2320 = 2422$ 50 ft obstacle at 5000
To correct for pressure altitude, we do the same thing at 6000 feet...
6,000 feet
$\left[ \left(1455 - 1345\right) \over 2 \right] + 1345 = 1400$ ground roll
$\left[ \left(2525 - 2320\right) \over 2 \right] + 2320 = 2732$ 50 ft obstacle
...and then interpolate again:
5,380 feet
$\left[ \left(1400 - 1265\right) * 0.380 \right] + 1265 = 1316$ ground roll
$\left[ \left(2732 - 2422\right) * 0.380 \right] + 2422 = 2540$ 50 ft obstacle

You are a champion!


text:
Why are westbound transatlantic routes located hundreds of km away from eastbound routes?

Are jet streams a net benefit in time and fuel savings? provides (basic) explanations about jet streams in the Pacific. However I'm looking for an answer focused on North Atlantic area, explaining the track system and the tracks location. In addition BA 185 route seems to be composed of two arcs, did they change their mind in flight because of the wind? or is this related to ETOPS or alternate airport at Santa Maria?

Looking at flights between NY and London (click to see route):

BA 185 (EGLL - KEWR)
United 941 (EGLL - KEWR)
United 16 (KEWR - EGLL)

The FlightAware anticipated routes are quite similar in both directions, but the actual routes for past flights are really remote from each other:

The two westbound routes are either 800 km north or 1,000 km south of the eastbound route (the dotted line shows the shortest path).

Why are the two westbound routes different and so remote from each other? Why this difference of about 1,800 km?
Why BA 185 route seems to be composed of two arcs?
Did the pilots changed their mind in flight because of the jet streams? Is it related to ETOPS constraints, or alternate airport at Santa Maria? or something else?

Possible duplicate of Are jet streams a net benefit in time and fuel savings?; The accepted answer explains it quite well.

Basically the same reason sailing ships follow the trade winds across the Atlantic. Of course, different winds at sea level than at cruise altitude.

Could it also have at least something to do with possible emergency landing sites?  I'd think the key sites in NE Canada could possibly guide the route coming westbound, and the sites in Ireland and coastal UK locations could direct the eastbound route???  And since the distance isn't as different as the map shows, it'd be worthwhile?

Cool :-)  What day(s) were the trips from, so I can pull up an upper air map?

The reason this is done is due to the winds aloft. The Jet Stream is a powerful current of air that blows in a west to east direction. Airplanes crossing the Atlantic from west to east take advantage of the jet stream to get there faster and save fuel, so a course is chosen to stay in the stream as much as possible. Airplanes going from east to west will be slowed by the jet stream, using more fuel and taking longer to get across, so a course is chosen to avoid the strongest winds.
The difference can be very significant, I've been on flights from NYC to London which take 6h 30m, and ones from London to NYC that take almost 9 hours.

I imagine both routes are approximately great circle arcs, meaning that the actual distance wouldn't be much different, even though they are so far apart. Worth mentioning if it's true.

@mins Actually I guess you can't unless the two positions are on opposite sides of the sphere. Not sure what I was thinking there.

@DCShannon still, the northerly route isn't as long as the map makes it look.

@DCShannon, there is only one great circle that connects any two given points on the globe.  If the route from here to there is different from the route from there to here, then at most, one of the two can be a great circle.

@jameslarge That is not true as DCShannon points out in the comment just below his original one. Assuming you're modelling the globe as a sphere any two points which are polar opposites get an infinite number of great circles. If you are not modelling the earth as a sphere you don't really get a great circle per se but you still get at least two distinct shortest paths for some points.

Wikipedia article about North Atlantic Tracks says it is to avoid jet stream winds (when flying west) or use it (when flying east):

They are aligned in such a way as to minimize any head winds and maximize tail winds impact on the aircraft. This results in much more efficiency by reducing fuel burn and flight time. To make such efficiencies possible, the routes are created twice daily to take account of the shifting of the winds aloft and the principal traffic flow

Article links to daily map of tracks, complete with the position of the jet stream.
Jet stream winds have average speed 100 mph (160 km/h) so they are significant.

authors: Юлия Федорова


text:
What happens aerodynamically when we trim an aircraft?
I'm trying to understand stability & control. Please correct me if I'm mistaken.
An aircraft will have longitudinal stability if the aerodynamic center (AC) is behind the center of gravity (CG). The AC is a point, where all the changes in the magnitude of the lift effectively take place.
The aircraft will be trimmed if the sum of the moments equals zero.
Question: is the center of pressure (CP) of the total aircraft (wing's lift and tailplane's aerodynamic force) or the AC, that must be positioned to coincide with the CG in order to trim the aircraft?
Do we change the location of the AC when we deflect the elevator?

It depends on the aircraft in question, the arrangement of the lifting surfaces, and how they are designed to ensure both static and dynamic stability.
For a simple case, well consider a traditionally designed airplane with a wing set placed approx midway along a fuselage with a horizontal tailplane at the empennage.
The design provides both good static stability and good dynamic stability.
It does so by placing the center of gravity (CG) ahead of the center of lift (CP) for the main wing and causing the tailplane to produce a downwards lifting force to counterbalance the moment caused by the distance between the CG and the CP.  While this is not as efficient a lifting arrangement as other designs like a canard foreplane, it forms a naturally stable platform in terms of longitudinal stability.  The airplane will remain stable only if these two moments cancel each other out causing the Net moment about the CG to be zero AND if the design naturally returns to a zero moment about the CG when dynamically changing speeds and attitudes.
For trimming, here we will only consider pitch trim for longitudinal stability.  An airplane trimmed for cruising at a specific airspeed and a particular altitude will have a net longitudinal moment about the CG of zero.  If the airplane increases its airspeed, the tailplane arrangement is designed to increase its downward force, creating a nose up pitch moment.  This in turn decreases the airspeed causing the the downward tailplane lift to decrease again, resulting in the nose pitching forward again until an equilibrium of forces and moments is again attained, which will be at the original airspeed for which the plane was trimmed for.  To maintain straight and level flight at a higher airspeed the trimming action will decrease the amount of lift that tailplane produces at the new selected airspeed, again resulting in a net longitudinal moment about the CG to zero.
Longitudinal trimming can be accomplished several different ways the first of which is a design feature which can alter the angle of attack of the tailplane in flight to change the lifting force it creates.  This is common on jetliners and other large aircraft but not common on light aircraft due to the excess weight of the structure.  Another option for smaller aircraft is the installation of a movable servo or antiservo tab on the elevator surfaces.  These tabs force the elevators to a new neutral position, again altering the lifting force from the tailplane and creating a trimmed condition.
In regards to the original question the net CP, that is the combined location of both the center of pressure of the wings and the center of pressure of the tailplane will coincide with the CG on the longitudinal axis of a trimmed airplane.  If the airplane accelerates, the NET CP will move fwd of the CG, causing the airplane to nose up, without additional trimming to bring it back.  Similarly, when the airplane decelerates, the NET CP will move aft of CG, causing the aircraft to nose down without additional trimming.

Net CP seems easier to understand when elevator is deflected down, increasing lift in back and dropping the nose.  A trim with CG forward of wing CP is elevator up.  Would not accelerating from trim bring nose up?  Just checking, answer made sense to me except last part.

the NET CP will move forward of the CG, causing the airplane to nose down Is this sentence correct? Sounds backwards.

The problem is fixed

No, looks like not fixed, the last paragraph is still contradictory. And anyway, this is not quite a correct explanation (why do you think the net CP would move with speed?) See footnote to my answer.

I'll offer a simpler and more direct answer: in a trimmed condition, it is a total CP that must coincide with the CG, and this is essentially by definition, and it doesn't "depend on the aircraft in question".
(Here for simplicity we restrict ourselves to the pitch motion and ignore possible effect of thrust and drag, which line may not exactly pass through CG and which moment will then need to be compensated).
In the same condition, AC will be behind CG (and CP) for a statically stable (in pitch) airplane. Imagine the airplane is disturbed and pitches up (or experiences an updraft; the fact is, its AoA temporarily increases). The extra lift due to increased AoA is applied at AC (now by the definition of AC), and since AC is behind CG, it creates a pitch-down moment which returns the aircraft to the original AoA until this extra lift is eliminated and everything returns to the former balance. This is the definition of (static) stability.(*)
From this it follows - and it is important to realise it clearly - that an airplane is trimmed for a certain AoA. Not the airspeed, not the pitch. At a given trim setting (for a steady level flight), you can fly at a higher speed and higher load, for example, in a turn or spiral.
Another thing that may help to avoid confusion is to understand that AC is a very theoretical, abstract point. It is defined purely for the convenience of stability analysis, and defined such that it doesn't move (within reasonable AoA). So in flight you can't "position" it at will, just as well as in most cases you can't move CG much. In a sense, all control is done by shifting the CP (of the whole aircraft).
At the same time, CP and CG can be though of as "real" points where a known real force is applied (although both are also abstractions in reality). When you need a balance, i.e. lack of total moment, you want the lift and gravity to act at the same point. (Remember we neglected moments from other forces, which are often small).

(*) Longitudinal stability is often incorrectly explained through speed: the aircraft pitches up, loses speed, then "wants" to return to the trimmed speed by pitching down to accelerate. This is wrong; the pitch-down moment arises immediately as the AoA grows, much sooner than any appreciable change of speed (if any) happens. When flight dynamicists speak of longitudinal static stability - and that's exactly where the concept of AC appears - they really speak about AoA stability. Airspeed is not even a factor there (or rather, change of airspeed is not). It is this AoA stability that makes the aircraft flyable by humans.
When we disturb only airspeed, as in Carlo's answer, e.g. by increasing thrust, a different process is involved. First (ignoring some fine effects), lift starts to increase quickly (as square of speed). But this increase doesn't come at AC; remember AC is only about AoA! Because we maintain the same AoA (at least initially), you get proportional increase of lift on the wing and tail so that the total balance remains, and the lift increases at CP=CG. As a result, the airplane starts to accelerate upwards (but not 'climb' in a normal sense). Now this means decrease of AoA, and, apart from damping of lift itself, this triggers the normal AoA response, that is, the attempt to increase it back to the trimmed AoA, i.e. pitch up.
Note I didn't mention the tail downforce. It's not really a requirement. It's just an exaggerated way to ensure a rearward AC position. But you don't need to involve it to explain stability if you already defined AC. The downforce is just an implementation detail, as programmers say.

By definition net Cp must take into account the downforce of the tailplane, making it a required component of the net Cp.  Also, not only does the lift from the main wing increase with the square of the speed for a given AoA, but so does the tailplane force, altering the net moment about the CG, causing the nose to pitch up.

@Carlo, no. First, not the downforce, but just lift of the tailplane in general. It can be up or down, even for a statically stable airplane. Second, exactly because both the wing and the tail lift increases proportionally (say, doubles), the net moment doesn't change with airspeed alone. (Consider the classic example with uneven weight balances: if you double the weight on each arm, nothing will change). This is different to changes due to AoA: there lifts change disproportionally due to typically lower incidence of the tail; say, doubles on the wing and triples on the tail, and tail wins.

@Zeus Do you mean the other way (last sentence)? For example when the AoA increases a bit by disturbance, both lifts increase, but lift on wing must be more dramatic for the whole plane to pitch down a bit to restore the AoA.

@Hot.PxL Yes in absolute terms the wing lift will likely increase more. But it is moments that matter here: we are talking about angular motion. The wing lift has a much shorter arm to CG (with respect to which we calculate everything). If we had a balance initially, then we can simply look at the proportions by which forces change. And to have a stabilising moment, the tail lift must increase by a greater percentage than the wing's.

@Zeus Yes, I'm talking about moment (with respect to CG). When AoA increases a bit by disturbance (probably wind coming from below the airplane), the stability requires the plane to pitch down a bit to lower the AoA. For it to lower pitch, it must be the case that moment generated by the wing is greater (causing pitch down) than that of the tail (causing pitch up). Is my understanding correct?

@Hot, I think you're assuming the tail produces downforce and hence pitch-up, and in a disturbance the magnitude of the lifts increase. But it doesn't have to be like that. Or rather, if you just increase speed, all the forces will simply scale up, with little or no angular effect (see my reply to Carlo). But when you add a certain AoA, everything changes. Your tail downforce may well turn into 'upforce', while the wing lift (and its moment) may change by say 10%. The relation between these changes will determine stability.


text:
How do airlines pay for their planes?
Taking easyJet as an example, they have a fleet of over 200 A319's and A320's. At a price tag of around \$90m per plane, that equates to about \$18bn.
Given that their turnover for 2012/13 was around \$7bn, and their net profit was only \$660m, how exactly do they own so many planes? Are they financed long-term? Are they leased?
How are the deals structured for the big airlines when purchasing new planes?

Recommend to have a look on their annual report and financial statements

The overwhelming majority of major airlines do lease at least some of their aircraft. And most airlines who choose to buy do not end up paying that full price tag.
In the case of Easyjet, around half their fleet is leased (as of 2013). Leasing allows airlines with weak balance sheets or with poor future prospects to increase capacity without locking capital.
In fact, the largest aircraft owners are aircraft leasing companies. For example, GECAS (General Electric Capital Aviation Services), the largest aircraft lessor, currently owns around 1,700 aircraft, being operated by 230 airlines around the world.
There are other (not state-owned and generally low cost) airlines, such as Ryanair, who are fond of owing most of their aircraft. What Ryanair does is negotiate very good procurement deals with Boeing (by buying large packs of planes, all being B-737s), use them for around 5 years, and then sell them before their price has gone down too low. This takes a burden on their debt levels, but seems to work for them so far on the long run.
This aircraft resell value is an important aspect to consider when comparing leasing vs purchasing, as well as tax incentives derived from amortization.
Legacy carriers (many of them state owned), on the other hand, have a tendency to purchase most of their fleet, and keeping them around for the full lifespan of the machine (around 30 years).
The actual price a company pays for a given aircraft is not public (understandably) and most figures you would be able to find are related to "list prices". However, in 2005 Ryanair was forced to disclose facts about one of the massive (by the time) purchases of Boeing aircraft. Apparently, they paid less than half of the public list price.
More current data on the actual price airlines pay for Boeing planes can be found here, where Javier Irastorza analyses the discounts Boeing is applying to their aircraft based on a balance sheet assessment.

I seriously doubt ALL airlines lease all of part of their fleets.

Thanks for pointing that out, Certainly airlines using small Code A or B aircraft may not have access to the leasing market.

and not just that, same with many small government owned airlines in say Africa that use old aircraft purchased on the second hand market or old military transports (in fact in quite a few countries based on the Soviet model the national airline IS part of the country's air force, and is their transport arm, Aeroflot in the Soviet era was the like that).

Excellent answer, but 2nd to last paragraph seems to be missing part of a sentence - "This makes sens for them as they do not..."

Dan, sorry, that was a sentence that was more an opinion than a fact, so I chose to delete, just was not hard enough with the DEL key :)

jwenting, I changed my response based on your comment, please read the first sentence. I'm now not referring to the totality. In any case, any documented examples you could provide would be most welcome. For example, Aeroflot, which I consider to be a major airline, as of today has most of its fleet leased from various lessors.

Isn't there some government regulation which stipulates ownership of a minimum number of planes for any airline?

Interesting links on Boeing's (estimated) actual selling prices.

Much of course depends on the companies involved, but there's several common schemes in place:

Lease
Effectively like your lease car, you pay a fee per month or year over a contract period. At the end you either get a right to purchase the aircraft for a nominal price or it reverts to the lease company, or you extend the lease.
Purchase Outright
Yes, you can pay cash. Especially older companies may have the cash reserves to pay for new aircraft out of pocket.
Bank Loans
Either a bank or the manufacturer underwrites a loan with the aircraft as collateral. Pretty similar to a mortgage on your house
Trade in
Like buying a new car, you might get a fair price for that old aircraft you had sitting around
Subsidies
There's still a lot of national prestige involved in some countries with getting foreign airlines to fly your product. Some aircraft manufacturers as a result can deliver well under the list price to foreign airlines because such sales may be heavily subsidised by their governments.

@HorusKol: As far as I understand accounting, the cost appears at once when the payment is deducted from the balance and added to property and then it's accounting cost is gradually reduced in the property account and deducted from expenses (this is called amortisation)

@HorusKol: And it works the same way with about anything (including cars), just the term increases with lifespan and price of the item.

@HorusKol: Yes. Well, the expense is not really a loss, because the company shelled out the money, but it got the property that has (initially) the same value. As it pays off in the following years, it is gradually amortized. The law even defines minimum time for amortization, because otherwise the company could deduct more, declare no net profit and pay no taxes.

@HorusKol It will be there as a cash expense. BUT at the same time there will be a 90 million increase in the company assets listed as well, namely the aircraft. As a result, the total financial balance of the company doesn't change. In following years, that 90 million will be gradually reduced as the value of the aircraft goes down with increasing age. And then after 5, 10, 20 years when it is sold the assets will be reduced by the remaining value of the aircraft while cash reserves go up by the value it is sold for.

While scrolling through the prices of aircrafts which cost’s more than the GDP of many countries, it seems logical to take aircrafts on lease rather than see one’s balance sheets drooping health. Well, this might not be the only reason why many airlines pick leasing options.
By taking leased aircraft the airlines are able to maintain a fuel efficient and latest designs fleet, have better survival options at the time of geo – economic crises and traffic fluctuations. So this becomes a very company specific and situation specific decision.
Aircraft leasing market is a very capital intensive business with high risks and higher returns. Since last few years this market is expanding –with Asia- Pacific region fuelling this growth. Asia-Pacific region is becoming the hub of emerging economies -China, Hong Kong, Japan to name some. Among these China is moving fast and is growing its fleet at a humongous rate. Boeing projected in its 2014 annual report that China will buy about six thousand aircraft worth 870 billion U.S. dollars by 2033.
Multiple financing options, reduced lease and interest rates, bulk orders, growing Low Cost Carriers are all amplifying the growth of Global Aircraft Leasing Market


text:
Which source to use when reporting the current altitude to ATC?
While flying from KSNA to KVCB VFR at night, Norcal Approach asked for my current altitude.  I was somewhere between 12.0 and 12.5 depending on which reading I was going to use.  ATC had me confirm.  My GPS altitude was 12.5, my transponder was showing 12.3, and my altimeter indicated 12.0.  I am assuming, since altimeter settings are based on the barometric pressure on the ground, that the thinner air above 12,000 feet would make my altimeter the least reliable of the three and the GPS the most reliable.
When ATC asks, should I give them the altimeter indicated altitude?  Also, is the transponder altitude that I can see derived from secondary radar and isn't that what ATC is already seeing?

Ask yourself this - if ATC told you to fly a certain altitude, which instrument would you use?

When asked for altitude, you report the altimeter reading, utilizing the correct barometric pressure entered into the Kollsman window.
ATC separates traffic based upon indicated altitude.  The indicated altitude may include errors, such as the pressure ATC provides, but all aircraft in the area will presumably be using the same barometric pressure, and will have similar errors.
GPS altitude, which some call an absolute altitude, is an altitude relative to a model, normally the WGS84 elipsoid, and has it's own set of errors.
By convention, indicated altitude, based upon local pressure, is used for aircraft separation.

"all aircraft in the area will presumably be using the same barometric pressure" and if, for any reason, you're uncertain about the altimeter setting, you can always ask for the current setting. Either separately, or just report something like "maintaining XXXX feet, altimeter 29.89, N345"; if you've got the wrong setting, ATC would hopefully catch that and give you the correct setting.

Based upon local pressure below transition level; above transition altitude it is standard pressure (but then it will be called flight level rather than altitude; in the USA the transition level is uniformly 18,000 ft, so you won't be above it in a piston single, but in other countries it may be as low as 5,000 ft).

Good point on the other countries.  I have run into that but have a North American bias in my answer as the OP cited US locations.

You should generally report what you see on your altimeter (at least that is what i was always instructed to do) but this presumes that you are up to date on your local pressure settings (which you should be). This also insures proper separation in the airspace system.
FWIW Transponders generally report the altitude calibrated to a set 29.92 and then rectified as necessary on the receiving end.
GPS altitude is triangulated off the system and (as far as I know) is never used for compulsory reporting as it does not provide proper separation altitude.
One thing to consider is that ATC may have been asking you to verify your altitude i.e. that you were off your expected altitude potentially due to an incorrectly set altimeter, the proper phraseology should have been

Note that exact altitude or flight level means to the nearest 100 foot
  increment. Exact altitude or flight level reports on initial contact
  provide ATC with information required prior to using MODE C altitude
  information for separation purposes. At times, controllers will ask
  pilots to verify that altitude. The phraseology used will be:
"Verify at (altitude)."


text:
Can you identify this plane from 1920 and prior vintage?
My grand uncle, that is my grandfather's brother had a flight school and performed in barnstorming shows somewhere in the midwest and it's not a Jenny. His name was Ralph Bloxham and is seen with his flying cap on in the rear cockpit.
I've found a few planes with similar characteristics but none with them all. Can you identify the plane?

Single bay biplane.
N- type interplane struts.
Little to no dihedral.
Rear cockpit is just below the trailing edge of the top wing, no cutout in the top wing.
Dual (upper and lower wing) ailerons with push/pull link.
Non-radial (inline or V engine, likely water-cooled but possibly air-cooled such as a Franklin)
Two blade prop.
Bipod cabane struts at the front spar, single rear spar cabane strut.
Mild stagger of the top wing forward, Wire wheels (may be missing covers)
No brakes, interesting side panel on fuselage with a radiused front edge, rather long- coupled empennage which is not visible in the photo.
Fuselage is rather slab sided with pretty constant taper and is very slender aft of the cockpits.
Bottom of engine cowl tapers up toward the front, and no protruding push rods or exhaust manifolds visible.
Engine cover is not higher than the fuselage contour line (no hump)
Exhaust out the side or both sides, don't know but there's some semblance to the liberty engine.

The N strut design seems to be a rather uncommon trait as I search, most are wired only, and so is the completely enclosed long engine cowl/cover which rules out all radial or rotary types. Single bay rules out many. What do you think?

Welcome to aviation.SE!

I'm pretty sure that this is a Lincoln PT, part of the Lincoln Page PT series, although the PT/PT-K was in the late 20's and early 30's:

Source: Wikimedia Commons
You can read about it from EAA:

The Lincoln PT-K model was the next development of the Lincoln-Page PT, designed specifically as a flight training or low cost sport flying airplane. The “Page” had been dropped from the company name before the PT-K came out, so the “PT” stood for “Page Trainer”.
The PT-K was considered a well designed airplane from an aerodynamic standpoint in 1929, taking into account the mannerisms of a student pilot. The fuselage had a long moment arm for the tail surfaces, meant to desensitize the pitching and bucking of a nervous student pilot as he worked his way through the curriculum. In other words, the Lincoln was not too fussy and wouldn’t get all upset due to a little pilot error.
The PT-K was powered by a 100 hp Kinner K5, which could max out at 104 mph. It was no exceptional trick for the PT-K to perform a well rounded loop from level flight; there was really no need to dive and gain excessive speed to bring it around.
The structure of the PT series proved to be quite rugged. The original design had a split axle landing gear configuration with a shock absorbing bungee chord. The later design featured a stiff-legged rigid gear, coupled with low pressure airwheels. The low pressure airwheels had more than enough “give” in the sidewall so that no further shock absorption was needed.
The PT-K was available with a few options, including a metal propeller, a lightly framed fully enclosed canopy for cold weather flying, navigation lights, and engine starter. One other item available as an option was Bloxham “Safety Sticks”. This option allowed the instructor to disengage the student’s control stick if a student “froze” during the flight.

The aircraft in the picture is restored, but the strut matches especially around the cockpit.
The Lincoln PT was offered with a different V-8 engine, the Curtiss OX-5 but there is no evidence that one was ever sold. You sir may have that evidence as you don't seem to have the PT-K version, but the PT version.

+1 Looks like you found the right aircraft. There is other evidence for OX-5 powered PTs, however.

@PeterKämpf Maybe, I would question the Wikipedia source for that (it isn't cited), the picture you linked to seems to be an advertisement from Lincoln, it's possible that it is the same aircraft as the OP, or that it was converted to a -K before being sold. All the currently registered PT's are the -K model, which isn't to say that they weren't made/sold in the 30's, but there seems to be no surviving examples.

Thank you! It seems I came to the right place. I haven't heard of the LIncoln aircraft before. And the modification is actually known as the Bloxham Safety sticks? My uncle must have created that I imagine. Or he with help from his brother who was an engineer. I can see the struts are exactly the same and the long coupling was evident as I couldn't even see the empennage in the old photo. I see the powerplant is different. Thanks a bunch!

Yes, they were called "Bloxham Safety Sticks" sold by the Bloxham Aero Supply Company, however I can't find much information on that company or what became of it. They were actually marketed as "Bloxham Safe-T-Sticks"

OX-5 powered PT, so some were built.

There was also the Lincoln-Page LP-3.